# Exchanging limit and Gaussian integral

This is a physically motivated mathematical question. My goal is to compute the integral of the function $$f(x)=\exp\left(-\frac{x^2}{\Omega^2}+i\theta x\right)$$ in the case $$\Omega\to +\infty$$. This is a simple Gaussian integral, but I'm uncertain about one thing. If I take the limit before computing the integral, then $$\int_{-\infty}^{+\infty} dx \ \lim_{\Omega\to+\infty}f(x)=\int_{-\infty}^{+\infty} dx \ \exp(i\theta x)=\infty.$$ On the other hand, if I take it afterwards I get $$\lim_{\Omega\to +\infty} \int_{-\infty}^{+\infty} dx \ f(x)=\lim_{\Omega\to+\infty}\sqrt{\pi}\Omega \exp\left(-\frac{\Omega^2\theta^2}{4}\right)=0.$$ Now, I know there's a whole theory in mathematics that is about exchanging limits and integrals. But I'm a humble physicist and I care about consistency between the two results. If I know ahead of time that $$\Omega$$ is arbitrarily large, do I just do away with the squared term as if it never existed, and treat the finite case knowing that the infinite one cannot be derived as its limit?

• No. By taking the limit before the integral, you get $e^{i\theta x}$, which is simply not Lebesgue integrable. So, you just don’t write the symbol $\int_{-\infty}^{\infty}e^{i\theta x}\,dx$. Writing this symbol or saying this is equal to $\infty$ has as much meaning as the equation $\textit{sponge}\ddot{}\text{patrick}\ddot{\smile}\text{bob}=\text{mr krabs}$. So, you just don’t put the limit inside the integral. Once you learn about tempered distributions, you can, with a whole bunch of caveats about notational abuse, regard this as the Fourier transform of $1$, and say it equals $\delta$. Jul 7, 2023 at 8:32
• @ClaudeLeibovici What do you mean exactly? I know how to take the integral, there is a simple formula for that. Jul 7, 2023 at 8:43

In the present case, you miscalculated the first limit; indeed, one has by definition $$\int_\mathbb{R}\mathrm{d}x \lim_{\Omega\to+\infty} f(x) = \int_\mathbb{R}\mathrm{d}x\ e^{i\theta x} = 2\pi\delta(\theta).$$ As for the second limit, it vanishes unless $$\theta$$ is already zero, so that $$\lim_{\Omega\to +\infty} \int_\mathbb{R}\mathrm{d}x\ f(x) = \lim_{\Omega\to+\infty} \sqrt{\pi}\,\Omega\,e^{-\frac{1}{4}\Omega^2\theta^2} = \begin{cases} \infty &\mathrm{if}\; \theta = 0 \\ 0 &\mathrm{otherwise} \end{cases}$$ Then, you could show that this expression behaves as a Dirac delta by applying it to a test function $$-$$ but I'm a bit too lazy today to do it...
Nonetheless, this behaviour could have guessed from the start, since $$f(x)$$ corresponds to a normal distribution (up to normalization factor) with a shrinking standard deviation and the Dirac delta can be precisely modelled by normal law with a vanishing variance.